1

When we plot the attacker's computational power against the probability of double spending with a number of blocks equal to zero, we always get a chance of 100%.

This plot is based on the original white paper that models the probability of success as a Poisson distribution.

Is this correct?

Shouldn't the probability of success then be related to the computational power the attacker possesses?

  • Can you point to the specific calculation in the paper which you are talking about? I suspect it's essentially the trivial fact that an attacker who's 0 blocks behind is already caught up. – Nate Eldredge Sep 28 '17 at 2:48
  • @NateEldredge Yes, looking at the values: at q=0.1, z=0 P=1.0000000 and at q=0.3, z=0 P=1.0000000, it doesn't make sense to me that if there are zero blocks the attacker has 100% chance of winning the attack. Unless we assume that at zero blocks the transaction is incomplete/unconfirmed, we cannot say the probability of success is 100%. I even believe if the transaction is incomplete/unconfirmed, the probability of success is not 100%. – user1 Sep 28 '17 at 13:46
  • Which line on which page? – Nate Eldredge Sep 28 '17 at 13:58
  • These are the numerical results are on page 8. If you simulate the results yourself by modeling the probability of success as the Poisson distribution you'll get these values. – user1 Sep 28 '17 at 14:02
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It is implicitly assumed that the attacker performs a Finney attack, and hence gets a block for free. If the merchant accepts a 0-conf tx, the attacker will proceed immediately to release a double-spending block he has already computed, and thus succeed.

In any case, Satoshi's analysis is approximate, see https://bitcoil.co.il/Doublespend.pdf for a more accurate one (which makes a similar assumption).

0

The probability of success is related to the computational hashing power of the attacker in comparison to the total hashing power of the network.

If the attacker has 10% of the total hashing power (imagine one of the bigger mining pools) then he has 10% chance of getting the next block. And 10% chance of getting the following one; and 10% of getting the one after that.

So the chance that attacker will get the next block is

BLOCK NUMBER      CHANCE
1                 .1
2                 .01
3                 .001
4                 .0001
5                 .00001
6                 .000001

As you can see, after an hour the chance of an attacker maintaining the longest block is small (and extraordinarily expensive).

  • I'm looking for an answer from the probability side. I mean if you try to model this with the Poisson distribution or negative binomial distribution you dont get 10% with 1 block and so on. – user1 Sep 27 '17 at 19:55
  • This doesn't look correct to me. If I have 49.99999% of hashpower, it ought to be pretty much a coin flip for me to reverse a transaction 10 blocks back, not a 1 in 1000 chance. – Nick ODell Oct 27 '17 at 20:15
  • My example was with 10% of the hashing power. If you have 50% of the hashing power you're correct it's pretty much a coin toss that you would have to win several times in a row. It would be you versus the rest the world. And, if you were identified as the founder of a 51% attack you would probably have to get a lot more than 2 or 3 blocks ahead before people gave up. You would be putting in an awful lot of money and effort reversing one transaction. And, you would break the trust in BTC. – Mayo Oct 31 '17 at 12:54

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